Question

Coasting to a Stop Assume that the resistance encountered by a moving object is proportional to the object's velocity so that its velocity is \(v=v_{0} e^{-(k / m) t} .\) (a) Integrate the velocity function with respect to \(t\) to obtain the distance function \(s .\) Assume that \(s(0)=0\) and show that $$s(t)=\frac{v_{0} m}{k}\left(1-e^{-(k / m) t}\right).$$ (b) Show that the total coasting distance traveled by the object as it coasts to a complete stop is \(v_{0} m / k .\)

Short answer

The integrated form of the velocity function, which represents the distance as a function of time, is \(s(t) = \frac{v_{0}m}{k}\left(1 - e^{-(k/m)t}\right)\). The total coasting distance when the object comes to a stop is \(\frac{v_{0}m}{k}\).

Step-by-step solution

Integrate the velocity function to obtain the distance function

Begin with the given velocity function \(v = v_{0} e^{-(k / m) t}\). Note that this is the derivative of the distance function with respect to time. To reverse engineer the distance function, integrate the velocity function with respect to time. \[ \int v \, dt = \int v_{0} e^{-(k / m) t} \, dt = v_{0} \cdot \left( \frac{-m}{k} \right) e^{-(k / m) t} + C \] Now it's necessary to determine the constant of integration C. The given problem states that the initial distance is zero, or \(s(0) = 0\), which can be substituted into the integrated function to solve for C.

Solve for the integration constant C and obtain the distance function

Substitute \(s(0) = 0\) into the integrated function and equate this to zero to solve for C: \[0 = v_{0} \cdot \left( \frac{-m}{k} \right) e^{-(k / m) * 0} + C = \frac{-v_{0}m}{k} + C\] Solving for C gives \(C = \frac{v_{0}m}{k}\). Now, substitute C back into the integrated function to obtain the distance function: \[ s(t) = v_{0} \cdot \left( \frac{-m}{k} \right) e^{-(k / m) t} + \frac{v_{0}m}{k} = \frac{v_{0}m}{k}\left(1 - e^{-(k/m)t}\right)\] This is the desired expression for the distance function.

Determine the total coasting distance

The total coasting distance is attained when the object comes to a stop, that is, when \(v = 0\). Per the provided velocity function, \(v = 0\) when \(t\) approaches infinity. Substitute \(t = \infty\) into the distance function obtained in step 2 to determine the total coasting distance: \[s(\infty) = \frac{v_{0}m}{k}\left(1 - e^{-(k/m)*\infty}\right) = \frac{v_{0}m}{k} \] Since \(e^{-(k/m)*\infty}\) tends to 0 as \(t\) approaches infinity. This shows that the total coasting distance is \(\frac{v_{0}m}{k}\), as needed.

Key concepts

Calculus

Calculus is a mathematical toolbox that allows us to analyze changes and model dynamic systems. It is divided mainly into two branches: differential calculus and integral calculus. In the context of our exercise, calculus helps us to understand how an object's velocity changes over time and how to find the total distance traveled.

Differential calculus focuses on the rate of change, which is often represented by derivatives. The velocity of an object, for example, is the derivative of its position with respect to time. Conversely, integral calculus is used to accumulate these changes, allowing us to move from velocity (rate of change of distance) to displacement or distance traveled. In our problem, we start with the velocity function and use integration to find the corresponding distance function.

Exponential Decay

Exponential decay describes a process where a quantity decreases at a rate proportional to its current value. This type of decay is common in many natural phenomena, such as radioactive decay in physics or depreciation in finance.

The general form of an exponential decay function can be described by the equation \(( y = y_0 e^{-kt} \)), where \(( y_0 \) is the initial quantity, \(( k \) is a positive constant that determines the rate of decay, and \(( t \) is time. In our velocity function \(( v = v_0 e^{-(k/m) t} \)), we see this pattern, indicating that as time increases, the velocity of our moving object decreases exponentially. The constant \(( k/m \) dictates how quickly the velocity decreases, with larger values leading to faster deceleration.

Definite Integral

A definite integral computes the accumulation of quantities, such as the total distance traveled by an object, over a given interval. The process involves finding the area under the curve of a function on a graph within defined limits. When we integrate a velocity function over time, we essentially sum up all the infinitesimal distances covered during each moment in time, which gives us the total distance.

With reference to our problem, integrating the velocity function \(( v(t) \) from time \((0\) to any time \((t\) gives us the exact distance function, denoted by \((s(t)\)). This shows the distance the object has traveled at any time \((t\). The 'definite' aspect of this integral comes into play when we examine specific time intervals, such as from \((0\) to \((t\), or \((0\) to infinity to find the total coasting distance.

Constant of Integration

When we integrate a function, we often add a constant of integration, denoted usually by \((C\)). This arises because integration is the reverse process of differentiation, and when differentiating, adding a constant to a function does not change its derivative. Therefore, when we integrate, we must account for all the possible constants that could have been there before differentiation.

In practical problems, like the one we're tackling, initial conditions allow us to find this constant's exact value. The problem statement indicates that the distance at time \((0\) is zero, \((s(0) = 0\)), so we use this to solve for the constant \((C\)). The constant of integration gives us a complete distance function that appropriately represents the physical situation we are analyzing. Through this, we can accurately calculate how far the object will travel at any given time.